This was done from using the method of synthetic division. I attached the image. Hope this helps
1. the image is multiplied by the scale factor. if the scale factor is 2, the image is twice the size of the preimage. if the scale factor is ⅓, the image is one third the size. only a scale factor of 1 preserves congruency
The new image often has a ' next to it if this helps
![\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta) \\\\\\ tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\ -----------------------------\\\\ 2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)} \\\\\\ 2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A) \\\\\\ 2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2](https://tex.z-dn.net/?f=%5Cbf%20sin%5E2%28%5Ctheta%29%2Bcos%5E2%28%5Ctheta%29%3D1%5Cimplies%20cos%5E2%28%5Ctheta%29%3D1-sin%5E2%28%5Ctheta%29%0A%5C%5C%5C%5C%5C%5C%0Atan%28%5Ctheta%29%3D%5Ccfrac%7Bsin%28%5Ctheta%29%7D%7Bcos%28%5Ctheta%29%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%0A2cos%28A%29%3D3tan%28A%29%5Cimplies%202cos%28A%29%3D3%5Ccfrac%7Bsin%28A%29%7D%7Bcos%28A%29%7D%0A%5C%5C%5C%5C%5C%5C%0A2cos%5E2%28A%29%3D3sin%28A%29%5Cimplies%202%5B1-sin%5E2%28A%29%5D%3D3sin%28A%29%0A%5C%5C%5C%5C%5C%5C%0A2-2sin%5E2%28A%29%3D3sin%28A%29%5Cimplies%202sin%5E2%28A%29%2B3sin%28A%29-2)
![\bf \\\\\\ 0=[2sin(A)-1][sin(A)+2]\implies \begin{cases} 0=2sin(A)-1\\ 1=2sin(A)\\ \frac{1}{2}=sin(A)\\\\ sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\ \frac{\pi }{6},\frac{5\pi }{6}\\ ----------\\ 0=sin(A)+2\\ -2=sin(A) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5C%5C%5C%5C%5C%5C%0A0%3D%5B2sin%28A%29-1%5D%5Bsin%28A%29%2B2%5D%5Cimplies%20%0A%5Cbegin%7Bcases%7D%0A0%3D2sin%28A%29-1%5C%5C%0A1%3D2sin%28A%29%5C%5C%0A%5Cfrac%7B1%7D%7B2%7D%3Dsin%28A%29%5C%5C%5C%5C%0Asin%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%3D%5Cmeasuredangle%20A%5C%5C%5C%5C%0A%5Cfrac%7B%5Cpi%20%7D%7B6%7D%2C%5Cfrac%7B5%5Cpi%20%7D%7B6%7D%5C%5C%0A----------%5C%5C%0A0%3Dsin%28A%29%2B2%5C%5C%0A-2%3Dsin%28A%29%0A%5Cend%7Bcases%7D)
now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1
now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine
so, only the first case are the valid angles for A